Differential Calculus I PDF

Lecture notes 3.2 Differential Calculus I 3.2 Differential Calculus I 1 How fast is Usain Bolt? Usain St Leo Bolt is a Jamaican sprinter, six-time Olympic champion and world record holder in the 4 x 100 metre relay, in the 10...

Lecture notes 3.2 Differential Calculus I 3.2 Differential Calculus I 1 How fast is Usain Bolt? Usain St Leo Bolt is a Jamaican sprinter, six-time Olympic champion and world record holder in the 4 x 100 metre relay, in the 100 and 200 metre races. With a current world record time of 9.58 seconds, set on 16 August 2009 in the final of the World Championships in Athletics in Berlin, he is the only person to run the 100 metres in less than 9.6 seconds to date. This results in an average speed v of: a) The table shows the running times for 10 m sections of his 100 m record run. Enter the corresponding points in the coordinate system. Distance covered (in metres) 10 20 30 40 50 60 70 80 90 100 Intermediate time (in s) 1.98 2.88 3.78 4.64 5.47 6.29 7.1 7.92 8.75 9.58 Running time of the 10 m (in s) v of the 10 m (in m/s) Distance in m 100 90 80 70 60 50 40 30 20 10 0 Time in s 1 2 3 4 5 6 7 8 9 10 b) Explain that the average speed does not correspond to the maximum speed. c) Calculate his speed in the individual sections. Enter the values in the table. KSWO 2024 1 Lecture notes 3.2 Differential Calculus I d) Determine the section in which Usain Bolt ran the fastest. Describe how this could also be determined using the graph. e) Explain to what extent you are satisfied with this answer to the question ”how fast is Usain Bold?”. If necessary, describe a method that could be used to determine the answer more precisely. In the example of Usain Bolt’s, you have drawn and y f (x) analysed the function s(t), i.e. the distance covered as a function of time. Now let’s take a look at an arbitrary P1 function f (x) and do something similar. Given are 2 points P0 (x0 |f (x0 )) and P1 (x1 |f (x1 )) of the function f (x). The ∆y f (x1 ) − f (x0 ) quotient = describes P0 ∆x x1 − x0 x 2 Difference quotient For a given function f (defined on the intervall I ⊂ R) y f (x) and two different points x0 , x1 ∈ I, the term ∆y f (x1 ) − f (x0 ) P1 = ∆x x1 − x0 f (x1 ) is called difference quotient. ∆y It specifies the gradient of the secant through the P0 points (x0 , f (x0 )) and (x1 , f (x1 )). f (x0 ) ∆x Alternatively, x1 is often denoted as x0 + ∆x or as x0 + h. The difference quotient then is: x0 x1 x ∆y f (x0 + ∆x) − f (x0 ) f (x0 + h) − f (x0 ) = = ∆x ∆x h KSWO 2024 2 Lecture notes 3.2 Differential Calculus I Example 1: Given is the function f (x) = − 13 x3 , x ∈ R. Calculate the average rate of change m of f in the interval [6; 9]. Solution: ∆y f (x1 ) − f (x0 ) f (9) − f (6) − 1 (9)3 − (− 13 (6)3 ) −243 + 72 m= = = = 3 = = −57 ∆x x1 − x0 9−6 3 3 Example 2: Notes: KSWO 2024 3 Lecture notes 3.2 Differential Calculus I 3 Example Meteor Crater There is a meteorite impact crater in Arizona, the so-called Meteor Crater or Barringer Crater. The crater has a diameter of up to 1200 metres and a depth of 180 metres. At a very shallow point, the partial cross-section of the crater up to the rim can be described by the function k(x) = 0.002x2 for 0 ≤ x ≤ 300. There is a vehicle in the crater that can manage a gradient of up to 115%. Can the vehicle reach the crater rim and drive out of the crater? Tip: What does 115% gradient mean? If a gradient is indicated, e.g. on a road sign, the percentage means a change in height of 20 metres for every 100 m horizontal distance. You can find the exact information in the following figure: KSWO 2024 4 Lecture notes 3.2 Differential Calculus I 4 Derivative at a point The difference quotient only indicates the average rate of change of the function in the interval ∆x. However, we are often interested in the instantaneous rate of change. y f (x) To get to grips with this formally and mathemati- cally, we use an idea that is as simple as it is con- P1 0 vincing: We let ∆x become smaller and smaller and f (x1 ) x → ∆ ine examine the behaviour of the difference quotient for ntl a ∆y ∆x → 0 or x1 → x0 or h → 0. Geometrically, the sec secant through P0 and P1 (and the corresponding P0 ∆x f (x0 ) gradient of the secant line) then becomes a tangent tangent line through P0 (with the corresponding gradient of the ∆x → 0 tangent line). x0 x1 x The gradient of a tangent line We now analyse the gradient of a secant line from a fixed point on a curve over successively smaller intervals. How can the gradient of a tangent to a curve be found at any point? For example, the point P0 (1, 1) lies on the graph of the function f (x) = x2. What is the gradient of the tangent at this point? We could proceed as follows: 1. We assume that P1 lies on f (x) and has the coordinates P1 (x, x2 ). ∆y f (x1 )−f (x0 ) 2. The gradient of the secant line P0 P1 is: m = ∆x = x1 −x0 = x2 −1 x−1 3. We now calculate the gradients of some secant lines if P1 approaches P0 : f (x) = x2 x Point P1 Gradient of P0 P1 y 5 (5|25) 25 4 (4|16) 16 3 (3|9) 2 (2|4) 9 1.5 (1.5|2.25) 4 2.25 1.1 (1.1|1.21) 1 P0 1.01 (1.01|1.0201) x −1 0 1 2 3 4 5 4. Conclusion: KSWO 2024 5 Lecture notes 3.2 Differential Calculus I The above illustration therefore shows that if x approaches 1, the gradient of the secant line approaches the gradient of the tangent line at the value x = 1. We describe this process as follows: ! x2 − 1 x2 − 1 if x → 1, then →2 or lim =2 x−1 x→1 x−1 f (x1 )−f (x0 ) In general, the following applies: if x1 −x0 can be approximated arbitrarily close to a real number f (x1 )−f (x0 ) m by an arbitrarily close approximation of x1 to x0 , we say that the difference quotient x1 −x0 has the limit m. We write f (x1 ) − f (x0 ) lim = m. x1 →x0 x1 − x0 m is then the gradient of the tangent line to the curve at x = x0. Definition: If the limit df f (x1 ) − f (x0 ) f ′ (x0 ) = (x0 ) = x lim →x or dx 1 x ̸=x 0 x1 − x0 1 0 df f (x0 + h) − f (x0 ) f ′ (x0 ) = (x0 ) = lim or dx h→0 h df f (x 0 + ∆x) − f (x0 ) f ′ (x0 ) = (x0 ) = lim dx ∆x→0 ∆x exists, it is called the derivative of f (x) at x0 and is denoted by f ′ (x0 ) ( ”f prime of x0 ”) or df dx (x0 ) (”the derivative of f (x) with respect to x” or ”d f (x) d x”). The function is then called differentiable at x0. (In order to keep the notation simple, we will omit the addition x1 ̸= x0 in future). The 1st derivative of the function at x0 measures the gradient of the tangent line to the graph at the point (x0 , f (x0 )), i.e. the instantaneous rate of change. Example 3: Determine the gradient of the tangent line to the curve y = x2 at the point (2, 4). Solution: f (x1 ) − f (x0 ) x1 =2+h f (2 + h) − f (2) (2 + h)2 − (2)2 x0 = 2, m = lim = lim = lim x1 →x0 x1 − x0 h→0 2+h−2 h→0 h 2 4 + 4h + h − 4 4h + h 2 h(4 + h) = lim = lim = lim = lim (4 + h) = 4 h→0 h h→0 h h→0 h h→0 KSWO 2024 6 Lecture notes 3.2 Differential Calculus I Exercises: 1. Determine the gradient of the tangent line of y = 3 − x − x2 at x = −1. 9 2. Find te derivative of f (x) = x at x = 2. √ 3. Determine the derivative of f (x) = x at x = 9. 4. Find functions that are not differentiable at a point x0. KSWO 2024 7 Lecture notes 3.2 Differential Calculus I 5 The derivative of a function For a non-linear function y = f (x), the gradient of the curve changes continuously. For practical reasons, it is helpful to determine a ”gradient function” so that the gradient can be found at any point x = x0 can be found, if x is replaced with x0 in the gradient function. Definition: A function is differentiable if it is differentiable over its entire domain D. If f is differentiable, the function that assigns the value f ′ (x) to each point x ∈ D is called 1st derivative of the function f with respect to x, and it is denoted by df f′ oder. dx If you calculate the 1st derivative of a given function, you differentiate the function. This is df f (x + h) − f (x) written as follows: f ′ (x) = (x) = lim dx h→0 h Since the 1st derivative of a given function f (x) is itself also a function (f ′ (x)), you can also differentiate this function again and thus obtain the 2nd derivative of the function f : d2 f f ′′ or f (2) or d2 x. Summary: Example 4: KSWO 2024 8 Lecture notes 3.2 Differential Calculus I Exercises: 1.) Find the derivative of f (x) = x2. 2.) Find the derivative of f (x) = x1. √ 3.) Find the derivative of f (x) = x. 4.) Differentiate ”graphically”: https://www.geogebra.org/m/nUcgDV3m Write down your strategy for drawing the derivative of a function. KSWO 2024 9 Lecture notes 3.2 Differential Calculus I 6 Some rules for derivatives Complete the table. Find a formula for f ′ (x) if f (x) = xn , where n ∈ Q. Power rule: Function f (x) Derivative f ′ (x) x x2 x3 x4 x−1 = 1 x Proof: f (x) = xn ⇒ x−2 = 1 (x + h)n − xn x2 f ′ (x) = lim h→0 h n x + nhx n−1 + (poly. in h with exp. ≥ 2) − xn x−3 = 1 x3 = lim h→0 h √ h(nxn−1 + (poly. in h with exp. ≥ 1)) x1/2 = x = lim h→0 h n−1 = lim (nx + (poly. in h with exp. ≥ 1) = nxn−1 x−1/2 = √1 x h→0 Constant rule: f (x) = c Constant multiple rule: f (x) = c · u(x) Proof: Sum or difference rule: f (x) = u(x) ± v(x) Proof: Example 5: We can derive polynomials very easily using these derivation rules: f (x) = 3x4 + 2x3 − 5x2 + 7x + 6 ⇒ f ′ (x) = 3 4x3 + 2 3x2 − 5(2x) + 7(1) + 0 = 12x3 + 6x2 − 10x + 7

KSWO 2024 10 Lecture notes 3.2 Differential Calculus I Exercises: 1. Find f ′ (x) if f (x): a) 5x3 + 6x2 − 3x + 2 b) 7x − 4 x + 3 x3 2. Determine the derivative of f (x) = x2 − x4. Then find the gradient of the tangent line to the curve at x = 2. 7 Equations of tangent and normal lines The normal line at a point on a curve is the line perpendicular to the tangent line at that point. If the tangent line has the gradient mt , the normal line has the gradient mn = Example 6: Given are the function h(x) = 15 x2 and the point P (3| 95 ) on the graph of h(x). Determine y the equations of the tangent line t and the normal 4 h(x) line n to the graph of h(x) at the point P. 3 Instructions: 2 P 1. t(x) = mt x + qt , determine mt = h′ (3) 2. Insert P in t(x) and solve the equation for qt. 1 ⇒ t(x) x 3. n(x) = mn x + qn , determine mn = − m1t −2 −1 0 1 2 3 4 4. Insert P in n(x) and solve the equation for qn. −1 ⇒ n(x) KSWO 2024 11 Lecture notes 3.2 Differential Calculus I ⋆ Find a general formula for the tangent line t to the curve f (x) at the point A(a|f (a)). t(x) =... Exercises: 1) Find the equation a) of the tangent line and b) of the normal line to f (x) = x2 + 1 at x = 1. 2) Determine the horizontal tangent lines to f (x) = x3 − 12x + 2. The angle between the graphs of the functions f (x) and g(x) is the (smaller) angle between the tangent lines at the point of intersection S(xs |ys ) (where xs is found by equalizing f (xs ) = g(xs ) and solving for xs. Then ys = f (xs ) = g(xs )). φ = tan−1 (f ′ (xs )) − tan−1 (g ′ (xs )) , if φ > 90◦ : φ′ = 180◦ − φ f ′ (xs ) − g ′ (xs ) or tan(φ) = 1 + f ′ (xs ) · g ′ (xs ) 8 The chain rule
4 √ our goal is to find a rule to calculate the derivative of composite functions like e.g. x2 + 3x , 2 − 3x, 1. What these functions have in common is that they are made up of two simpler functions: x − x2 y = f (x) = u(v(x)).
4 ◦ y = x2 + 3x ⇒ u = v4 and v = x2 + 3x. √ ◦ y = 2 − 3x ⇒u= and v=. 1 ◦ y= ⇒u= and v =. x − x2 KSWO 2024 12 Lecture notes 3.2 Differential Calculus I Exercise: 1. Expand f (x) = (x2 + 3x)2 Differentiate each term to find the derivative of f (x). 2. f (x) = (x2 + 3x)2 is a composite function. Determine the outside function u and the inside function v? 3. Calculate the derivatives du dv and dv dx and find u′ (x) and v ′ (x) (hence get rid of v). Can you find a rule to calculate f ′ (x) using u and v? √ 4. Use this rule to calculate the derivative of f (x) = 7x5 + 2x The rule found above is called "the chain rule": dy The chain rule : If f (x) = u(v(x)) then f ′ (x) = or f ′ (x) = dx = du dv dv dx The derivative of a composite function is the derivative of the outside function with respect to the inside function (inside function remains the same), multiplied by the derivative of the inside function with respect to x. If the composite function is of the form y = (v(x))n , the chain rule is particularly simple: du dv dy u = vn → = nv n−1 , = v ′ (x) ⇒ = n(v(x))n−1 · v ′ (x) dv dx dx Example 7: KSWO 2024 13 Lecture notes 3.2 Differential Calculus I Exercises 4 Calculate f ′ (x) if: a) f (x) = (x2 − 2x)4 b) f (x) = √ 1 − 2x KSWO 2024 14 Lecture notes 3.2 Differential Calculus I 9 The product rule Prove by finding a counterexample that for a function f (x) = u(x) · v(x) the following is false: f ′ (x) = u′ (x) · v ′ (x). The product rule: If f (x) = u(x) · v(x) then f ′ (x) = u′ (x) · v(x) + u(x) · v ′ (x) The derivative of the product of two factors is the first factor times the derivative of the second factor plus the second factor times the derivative of the first factor. Proof: https://www.cuemath.com/calculus/product-rule/ 10 The quotient rule √ x2 + 1 x x3 Expressions like , , are also called quotients. Here we have a look at the 2x − 5 1 − 3x (x − x2 )4 derivative of the quotient of two functions: f (x) = u(x) v(x). u(x) u′ (x) · v(x) − u(x) · v ′ (x) The quotient rule: If f (x) = v(x) then f ′ (x) = (v(x))2 The derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the denominator squared. Proof: Note that u(x) = f (x)v(x). Using the product rule, this results in u′ (x) = f ′ (x)v(x) + f (x)v ′ (x) ⇒ u′ (x) − f (x)v ′ (x) = f ′ (x)v(x) u(x) ′ u′ (x)v(x) − u(x)v ′ (x) d.h., f ′ (x)v(x) = u′ (x) − v (x) = v(x) v(x) ′ ′ u (x)v(x) − u(x)v (x) ⇒ f ′ (x) = (v(x))2 KSWO 2024 15 Lecture notes 3.2 Differential Calculus I Example 8: KSWO 2024 16 Lecture notes 3.2 Differential Calculus I Exercises dy √
4 Calculate the derivative dx using the product rule. a) y = x(2x + 1)3 b) y = x2 x2 − 2x. √ dy x2 x Calculate the derivative using the quotient rule. a) y = b) y =. dx x2 −1 (1 − 2x)2 1st and 2nd derivative in physics (in german): https://lx3.mint-kolleg.kit.edu/onlinekursphysik/html/1.4.2/modstart.html KSWO 2024 17 Lecture notes 3.2 Differential Calculus I KSWO 2024 18